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For the given reaction, \mathrm{3 A+2 B \rightarrow C+D} , the experimental rate law is  \mathrm{R=K[A]^2[B]^1} . When the active mass of ' B ' is kept constant and ' A ' is tripled then, the rate of the reaction will:

Option: 1

Decrease by 3 times.


Option: 2

Increase by 9 times.


Option: 3

Increase by 3 times.


Option: 4

Unpredictable.


Answers (1)

best_answer

For the given reaction, the experimental rate law is, Rate =\mathrm{K}\mathrm{[\mathrm{A}]^2[\mathrm{~B}]^1}  When the concentration of ' A ' is tripled (3 A) and ' B ' is kept constant,
Rate  \mathrm{=K[3 A]^2[B]^1=9 K[A]^2[B]} 
Rate  \mathrm{=9 \times \mathrm{K}[\mathrm{A}]^2[\mathrm{~B}]} Thus, the rate of reaction is increased by 9 times.

Posted by

Gautam harsolia

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